A triangle with sides of 5, 12, and 13 has both an inscribed and a circumscribed circle.  What is the distance between the centers of those circles?
Explanation: Let $A$, $B$, and $C$ be the vertices of the triangle so that $AB = 5$, $AC = 12$, and $BC = 13$.  Let $I$ and $O$ be the incenter and circumcenter of triangle $ABC$, respectively.  Let the incircle of triangle $ABC$ be tangent to sides $BC$, $AC$, and $AB$ at $D$, $E$, and $F$, respectively.

[asy]
import geometry;

unitsize(0.4 cm);

pair A, B, C, D, E, F, I, O;

A = (5^2/13,5*12/13);
B = (0,0);
C = (13,0);
I = incenter(A,B,C);
D = (I + reflect(B,C)*(I))/2;
E = (I + reflect(C,A)*(I))/2;
F = (I + reflect(A,B)*(I))/2;
O = (B + C)/2;

draw(A--B--C--cycle);
draw(incircle(A,B,C));
draw(I--D);
draw(I--E);
draw(I--F);
draw(I--O);

label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
dot("$D$", D, S);
dot("$E$", E, NE);
dot("$F$", F, NW);
dot("$I$", I, N);
dot("$O$", O, S);
[/asy]

Since $\angle BAC = 90^\circ$, the circumcenter $O$ of triangle $ABC$ is the midpoint of hypotenuse $BC$.

Since $AE$ and $AF$ are tangents from $A$ to the same circle, $AE = AF$.  Let $x = AE = AF$.  Similarly, let $y = BD = BF$ and $z = CD = CE$.  Then $x + y = AF + BF = AB = 5$, $x + z = AE + CE = AC = 12$, $y + z = BD + CD = BC = 13$.  Solving this system of equations, we find $x = 2$, $y = 3$, and $z = 10$.  Then $DO = BO - BD = BC/2 - y = 13/2 - 3 = 7/2$.

The inradius $r$ of triangle $ABC$ is given by $r = K/s$, where $K$ is the area of triangle $ABC$, and $s$ is the semi-perimeter.  We see that $K = [ABC] = 1/2 \cdot AB \cdot AC = 1/2 \cdot 5 \cdot 12 = 30$, and $s = (AB + AC + BC)/2 = (5 + 12 + 13)/2 = 15$, so $r = 30/15 = 2$.

Hence, by the Pythagorean theorem on right triangle $IDO$, \[IO = \sqrt{ID^2 + DO^2} = \sqrt{2^2 + \left( \frac{7}{2} \right)^2} = \sqrt{\frac{65}{4}} = \boxed{\frac{\sqrt{65}}{2}}.\]